Question

A body weighs 20 N in a liquid whole relative density is 5 . In water it weighs 40 N . Find its weight in air?

Answer 1

Due to buoyancy force the weight is decreasing

So the weight of body = weight of water displaced by body in water.

So 40–20=20 N is weight of displaced water

20=mg

20=vdg

d=density of water=1 kg/cubm.

V=20/g

Answer 2

Explanation:

Density of a liquid = 5g/cm³ = 5000kg/m³

Buoyant force = Weight of the object in air - Weight of the object in a liquid

Vs (Volume of the solid) × density of the liquid ×g = Weight of the object in the air - 20N = 5000×10×Vs

Vs × density of the water ×g = Weight of the object in air - 40N = 1000×10×Vs

So, Weight of the object in air = 50000Vs + 20N

Weight of the object in air = 10000Vs + 40N

Equating both equations,

50000Vs+20N = 10000Vs+40N

==> 40000Vs = 20N

==> Vs = 1/2000 m³

Vs × density of the water ×g = Weight of the object in air -40 N

==> 1/2000 × 1000 × 10 = WOTOIA - 40N

==> 5 = Weight of the object in air -40 N

==> Weight of the object in air = 45N

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